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"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11 : Material Balance Problems involving Multiple Units"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 11.1 Page no. 311\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables \n",
"\n",
"w_A1 = 1 #concentration of A in 1\n",
"w_B2 = 1 # concentration of B in 2\n",
"w_A3 = 0.8 # concentration of A in 3\n",
"w_B3 = 0.2 # concentration of B in 3\n",
"w_C4 = 1 # concentration of C in 4\n",
"w_A5 = 0.571 #concentration of A in 5\n",
"w_B5 = 0.143 #concentration of B in 5\n",
"w_C5 = 0.286 #concentration of C in 5\n",
"w_D6 = 1 # concentration of D in 6\n",
"w_A7 = 0.714 # concentration of A in 7\n",
"w_B7 = 0.286 # concentration of B in 7\n",
"w_B8 = 0.333 #concentration of B in 8\n",
"w_C8 = .667 #concentration of C in 8\n",
"\n",
"us1 = 2 # Species involved in unit 1\n",
"us2 = 3 ; # Species involved in unit 2\n",
"us3 = 4 ; # Species involved in unit 3\n",
"\n",
"# Caculations \n",
"total_sp = us1+us2+us3 # Total species in system\n",
"\n",
"# Results\n",
"print 'Number of possible equations are 9, they are as follows- '\n",
"print ' Subsystem 1'\n",
"print ' A: F1*w_A1+F2*0 = F3*w_A3 (a)'\n",
"print ' B:F1*0 + F2*w_B2 = F3*w_B3 (b)'\n",
"print ' Subsystem 2'\n",
"print ' A: F3*w_A3+F4*0 = F5*w_A5 (c)'\n",
"print ' B:F3*w_B3 + F4*0 = F5*w_B5 (d)'\n",
"print ' C: F3*0+F4*w_C4 = F5*w_C5 (e)'\n",
"print ' Subsystem 3'\n",
"print ' A: F5*w_A5+F6*0 = F7*w_A7+F8*0 (f)'\n",
"print ' B:F5*w_B5 + F6*0 = F7*0+F8*w_B8 (g)'\n",
"print ' C: F5*w_C5+F6*0 = F7*0+F8*w_C8 (h)'\n",
"print ' D:F5*w_C5+F6*0 = F7*0+F8*w_C8 (i)'\n",
"print ' The above equations do not form a unique set'\n",
"\n",
"# By inspection we can see that only 7 equations are independent \n",
"#Independent Equations are: \n",
"# Subsystem 1\n",
"#A: F1*w_A1+F2*0 = F3*w_A3 (a)\n",
"#B:F1*0 + F2*w_B2 = F3*w_B3 (b)\n",
"#Subsystem 2\n",
"#A: F3*w_A3+F4*0 = F5*w_A5 (c)\n",
"# C: F3*0+F4*w_C4 = F5*w_C5 (e)\n",
"# Subsystem 3\n",
"#A: F5*w_A5+F6*0 = F7*w_A7+F8*0 (f)\n",
"#B:F5*w_B5 + F6*0 = F7*0+F8*w_B8 (g)\n",
"#D:F5*w_C5+F6*0 = F7*0+F8*w_C8 (i)\n",
"\n",
"print ' Number of independent equations are 7 '"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Number of possible equations are 9, they are as follows- \n",
" Subsystem 1\n",
" A: F1*w_A1+F2*0 = F3*w_A3 (a)\n",
" B:F1*0 + F2*w_B2 = F3*w_B3 (b)\n",
" Subsystem 2\n",
" A: F3*w_A3+F4*0 = F5*w_A5 (c)\n",
" B:F3*w_B3 + F4*0 = F5*w_B5 (d)\n",
" C: F3*0+F4*w_C4 = F5*w_C5 (e)\n",
" Subsystem 3\n",
" A: F5*w_A5+F6*0 = F7*w_A7+F8*0 (f)\n",
" B:F5*w_B5 + F6*0 = F7*0+F8*w_B8 (g)\n",
" C: F5*w_C5+F6*0 = F7*0+F8*w_C8 (h)\n",
" D:F5*w_C5+F6*0 = F7*0+F8*w_C8 (i)\n",
" The above equations do not form a unique set\n",
" Number of independent equations are 7 \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 11.2 Page no.315\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from numpy import matrix\n",
"\n",
"# Variables \n",
"\n",
"G = 1400 #[kg]\n",
"n_un = 16 # Number of unknowns in the given problem(excluding extent of reactions)\n",
"n_ie = 16 ; # Number of independent equations\n",
"d_o_f = n_un-n_ie # Number of degree of freedom\n",
"\n",
"print 'For unit 1 number of degree of freedom for the given system is %i .'%d_o_f\n",
"\n",
"o1_air = 0.995 ; # Mass fraction of air at out of unit 1 in A\n",
"i1_air = 0.95 ; # Mass fraction air at in of unit 1 in G\n",
"i1_wtr = 0.02; # Mass fraction water at in of unit 1 in G\n",
"F1_wtr = 0.81 ; # Mass fraction of water at out of unit 1 in F\n",
"o1_wtr = 0.005 ; # Mass fraction of water at out of unit 1 in A\n",
"o2_wtr = 0.96 ; # Mass fraction of water at out of unit 2 in B\n",
"o3_wtr = 0.01; # Mass fraction of water at out of unit 3 in D\n",
"i1_act = 0.03 ; # Mass fraction of acetone at in of unit 1 in G\n",
"F1_act = 0.19 ; # Mass fraction of acetone at out of unit 1 in F\n",
"o3_act = 0.99 ; # Mass fraction of acetone at out of unit 3 in D\n",
"o2_act = 0.04 ; # Mass fraction of acetone at out of unit 2 in B\n",
"\n",
"# Calculations \n",
"A = G*i1_air/o1_air ; #air-[kg]\n",
"F = G*i1_act/F1_act ; #[kg]\n",
"W = (F*F1_wtr+A*o1_wtr-G*i1_wtr)/1 #Pure water in -[kg]\n",
"\n",
"n_un = 9 ; # Number of unknowns in the given problem(excluding extent of reactions)\n",
"n_ie = 9 ; # Number of independent equations\n",
"d_o_f = n_un-n_ie \n",
"\n",
"print ' For unit 2 and 3 number of degree of freedom for the given system is %i .'%d_o_f\n",
"\n",
"a = matrix([[o3_act, o2_act],[o3_wtr, o2_wtr]]);\n",
"b = matrix([[F*F1_act],[F*F1_wtr]])\n",
"a = a.I\n",
"x = a*b \n",
"\n",
"# Results\n",
"print ' W-Pure water in to unit 1 - %.2f kg/hr'%W\n",
"print ' A-Air out of unit 1 - %.2f kg/hr'%A\n",
"print ' F-out of unit 1 - %.2f kg/hr'%F\n",
"print ' B-out of unit 2 - %.2f kg/hr'%x[1]\n",
"print ' D-out of unit 3 - %.2f kg/hr'%x[0]"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For unit 1 number of degree of freedom for the given system is 0 .\n",
" For unit 2 and 3 number of degree of freedom for the given system is 0 .\n",
" W-Pure water in to unit 1 - 157.74 kg/hr\n",
" A-Air out of unit 1 - 1336.68 kg/hr\n",
" F-out of unit 1 - 221.05 kg/hr\n",
" B-out of unit 2 - 186.15 kg/hr\n",
" D-out of unit 3 - 34.90 kg/hr\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 11.3 Page no. 318\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from numpy import matrix\n",
"\n",
"# Variables \n",
"P = 6205. #[lb mol/hr]\n",
"amt_F = 560. ; #[bbl]\n",
"C_F = 0.50 ; # [mol fraction]\n",
"H2_F = 0.47 ; #[mol fraction]\n",
"S_F = 0.03 ; #[mol fraction]\n",
"\n",
"CH4_G = 0.96 ; #[mol fraction]\n",
"C2H2_G = 0.02 ; #[mol fraction]\n",
"CO2_G = 0.02 ; #[mol fraction]\n",
"\n",
"O2_A = 0.21 ; #[mol fraction]\n",
"N2_A = 0.79 ; #[mol fraction]\n",
"\n",
"# Analysis of air into Oil furnace(A1)\n",
"O2_A1 = 0.20 ; #[mol fraction]\n",
"N2_A1 = 0.76 ; #[mol fraction]\n",
"CO2_A1 = 0.04 ; #[mol fraction]\n",
"\n",
"#Stack gas(P) analysis\n",
"N2_P = .8493 ; #[mol fraction]\n",
"O2_P = .0413 ; #[mol fraction]\n",
"SO2_P = .0010 ; # [mol fraction]\n",
"CO2_P = .1084 ; #[mol fraction]\n",
"\n",
"# Degree of freedom analysis \n",
"n_un = 5; \n",
"n_ie = 5 ;\n",
"d_o_f = n_un-n_ie; # Number of degree of freedom\n",
"print 'Number of degree of freedom for the given system is %i .'%d_o_f\n",
"\n",
"\n",
"# Calculations & Results\n",
"F = P* SO2_P/S_F ; # [lb mol/hr]\n",
"a = matrix([[2*CH4_G+C2H2_G, -1, 0, 0],[0, 0, N2_A, N2_A1],[CO2_G ,-.5, O2_A, O2_A1+CO2_A1], \n",
" [CH4_G+2*C2H2_G+CO2_G,0,0,CO2_A1]]);# matrix of coefficients\n",
"b = matrix([[-F*H2_F],[P*N2_P],[P*(O2_P+CO2_P+SO2_P)],[(P*CO2_P-F*C_F)]]); # matrix of constants\n",
"a = a.I\n",
"x = a*b \n",
"G = x[0]\n",
"m_F = 7.91\n",
"Fc = (F*m_F)/(7.578*42) # Fuel gas consumed -[bbl/hr]\n",
"time = amt_F/Fc ; # Time for which available fuel gas lasts-[hr]\n",
"\n",
" \n",
"print '(1) Fuel gas consumed(F) is %.2f bbl/hr .'%Fc\n",
"print '(2) Time for which available fuel gas lasts is %.0f hr .'%time\n",
"\n",
"# For increase in arsenic and mercury level\n",
"F_oil = Fc*42; #[gal/hr]\n",
"Em_ars2 = (3.96 *10**(-4))/1000.0 ; # [lb/gal]\n",
"Em_Hg2 = (5.92 *10**(-4))/1000.0 ; # [lb/gal]\n",
"ars_F = F_oil*Em_ars2 \n",
"Hg_F = F_oil*Em_Hg2 \n",
"G_gas = G*359 #[ft**3/hr]\n",
"Em_ars1 = (2.30 *10**(-4))/10**6 ; # [lb/ft**3]\n",
"Em_Hg1 = (1.34 *10**(-4))/10**6 ; # [lb/ft**3]\n",
"ars_G = G_gas*Em_ars1;\n",
"Hg_G = G_gas*Em_Hg1 ;\n",
"in_ars = ((ars_F-ars_G)/ars_G)*100 #[% increase in Arsenic emission]\n",
"in_Hg = ((Hg_F-Hg_G)/Hg_G)*100 #[% increase in Mercury emission]\n",
"\n",
"print '(3) Increase in Arsenic emission is %.1f %% .'%in_ars\n",
"print '(4) Increase in Mercury emission is %.1f %% .'%in_Hg"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Number of degree of freedom for the given system is 0 .\n",
"(1) Fuel gas consumed(F) is 5.14 bbl/hr .\n",
"(2) Time for which available fuel gas lasts is 109 hr .\n",
"(3) Increase in Arsenic emission is 107.4 % .\n",
"(4) Increase in Mercury emission is 432.3 % .\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 11.4 Page no. 322\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from numpy import matrix\n",
"\n",
"# Variables \n",
"M = 1000. ; #[lb]\n",
"F_s = 16/100. # Fraction of sugar in F\n",
"F_w = 25/100. # Fraction of water in F\n",
"F_p = 59/100. # Fraction of pulp in F\n",
"D_p = 80/100. # Fraction of pulp in D\n",
"E_s = 13/100. # Fraction of sugar in E\n",
"E_p = 14/100. # Fraction of pulp in E\n",
"G_p = 95/100. # Fraction of pulp in G\n",
"H_s = 15/100. # Fraction of sugar in H\n",
"K_s = 40/100. # Fraction of sugar in K\n",
"\n",
"# Calculations \n",
"K_w = 1 - K_s\n",
"K = M/K_s;\n",
"L = K_w*K;\n",
"\n",
"# For evaporator equations are \n",
"H_w = 1- H_s\n",
"H = K_s*K/H_s\n",
"J = H - K; \n",
"\n",
"# For screen equations are \n",
"E_w = 1 - (E_p + E_s) \n",
"\n",
"a1 = matrix([[1,-1],[E_p,-G_p]])\n",
"b1 = matrix([[H,],[0,]]) \n",
"a1 = a1.I\n",
"x1 = a1*b1\n",
"E = x1[0] \n",
"G = x1[1] \n",
"G_s = (E_s*E - H_s *H )/G\n",
"G_w = 1 -(G_s + G_p) \n",
"\n",
"a2 = matrix([[1,-1],[F_p,-D_p]]) # Matrix of coefficients of unknown\n",
"\n",
"F = 7818.93 \n",
"D = 1152.2634 \n",
"\n",
"D_s = (F_s*F - E_s*E )/D # By sugar balance\n",
"D_w = 1 -(D_s + D_p) # summation of wt. fraction is 1\n",
"\n",
"S_rec = M/(F*F_s) ; # Fraction of sugar recovered \n",
"\n",
"\n",
"# Results\n",
"print 'Flow streams and their respective compositions.'\n",
"print ' M = %.0f lb '%M\n",
"print ' Sugar: %.2f '%1\n",
"\n",
"print ' L = %.0f lb '%L\n",
"print ' Water: %.2f'%1\n",
"\n",
"print ' K = %.0f lb '%K\n",
"print ' Sugar: %.2f'%K_s\n",
"print ' Water: %.2f'%K_w\n",
"\n",
"print ' J = %.0f lb '%J\n",
"print ' Water: %.2f '%1\n",
"\n",
"print ' H = %.0f lb '%H\n",
"print ' Sugar: %.2f'%H_s\n",
"print ' Water: %.2f'%H_w\n",
"\n",
"print ' G = %.0f lb '%G\n",
"print ' Sugar: %.3f'%G_s\n",
"print ' Water: %.3f'%G_w\n",
"print ' Pulp : %.2f'%G_p\n",
"\n",
"print ' E = %.0f lb '%E\n",
"print ' Sugar: %.2f'%E_s\n",
"print ' Water: %.2f'%E_w\n",
"print ' Pulp : %.2f'%E_p\n",
"\n",
"print ' D = %.0f lb '%D\n",
"print ' Sugar: %.3f'%D_s\n",
"print ' Water: %.3f'%D_w\n",
"print ' Pulp : %.2f'%D_p\n",
"\n",
"print ' F = %.0f lb '%F\n",
"print ' Sugar: %.2f'%F_s\n",
"print ' Water: %.2f'%F_w\n",
"print ' Pulp : %.2f'%F_p"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Flow streams and their respective compositions.\n",
" M = 1000 lb \n",
" Sugar: 1.00 \n",
" L = 1500 lb \n",
" Water: 1.00\n",
" K = 2500 lb \n",
" Sugar: 0.40\n",
" Water: 0.60\n",
" J = 4167 lb \n",
" Water: 1.00 \n",
" H = 6667 lb \n",
" Sugar: 0.15\n",
" Water: 0.85\n",
" G = 1152 lb \n",
" Sugar: 0.014\n",
" Water: 0.036\n",
" Pulp : 0.95\n",
" E = 7819 lb \n",
" Sugar: 0.13\n",
" Water: 0.73\n",
" Pulp : 0.14\n",
" D = 1152 lb \n",
" Sugar: 0.204\n",
" Water: -0.004\n",
" Pulp : 0.80\n",
" F = 7819 lb \n",
" Sugar: 0.16\n",
" Water: 0.25\n",
" Pulp : 0.59\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 11.5 Page no.324\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables \n",
"F = 15. #[L/hr]\n",
"cs_in = 10. #Nutrient conc. input vessel - [g nutrient/L substrate]\n",
"V1 = 100. ; # [L]\n",
"V2 = 50. ; #[L]\n",
"Yxs = 0.2 ; # [cells/g]\n",
"umax = 0.4 ; #[hr** - 1]\n",
"Ks = 2. ; #[g/L] - Monod constant\n",
"\n",
"# Calculations\n",
"u1 = F/V1 #[hr** - 1] #[hr** - 1]\n",
"cs_out = (Ks * u1/umax)/(1 - (u1/umax)) \n",
"\n",
"x_out = Yxs * (cs_in - cs_out) #[g cells / L substrate]\n",
"\n",
"u2 = F/V2;\n",
"cs_out1 = (Ks * u2/umax)/(1 - (u2/umax)) #Nutrient conc. output vessel - [g nutrient/L substrate]\n",
"x_out1 = Yxs * (cs_in - cs_out1) #[g cells / L substrate]\n",
"\n",
"x_out2 = 1.73 # From eqn. (e),(f) and (g) - [g cells / L substrate]\n",
"\n",
"# Results\n",
"print 'g cells/L from option 1 is %.2f.'%x_out\n",
"print 'g cells/L from option 2 is %.2f.'%x_out2\n",
"print 'By comparing option 1 and option 2 the respective answers are essentially the same.'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"g cells/L from option 1 is 1.76.\n",
"g cells/L from option 2 is 1.73.\n",
"By comparing option 1 and option 2 the respective answers are essentially the same.\n"
]
}
],
"prompt_number": 13
},
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"cell_type": "code",
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